Question: $\dfrac{d}{dx}[-4e^x-\text{sin}(x)-9]=$
Solution: Recall that ${\dfrac{d}{dx}[e^x]=e^x}$ and ${\dfrac{d}{dx}[\text{sin}(x)]=\text{cos}(x)}$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[-4e^x-\text{sin}(x)-9] \\\\ &=-4{\dfrac{d}{dx}[e^x]}-{\dfrac{d}{dx}[\text{sin}(x)]}-\dfrac{d}{dx}[9] \\\\ &=-4{e^x}-{\text{cos}(x)}-0 \\\\ &=-4e^x-\text{cos}(x) \end{aligned}$ In conclusion, $\dfrac{d}{dx}[-4e^x-\text{sin}(x)-9]=-4e^x-\text{cos}(x)$